Question

What is the `pH` of `1MNaPO_(4)` solution at `25^(@)C`?
`PO_(4)^(3-)+H_(2)OhArrHPO_(4)^(2-)+OH^(-),K_(b)=2.4xx10^(-2)`
Assume no hydrolysis of `HPO_(4)^(2-)` ions.

Answer 1

`K_(a)=(x^(2))/(1-x)=2.4xx10^(-2)`
`x^(2)+(2.4xx10^(-2))xx-2.4xx10^(-2)=0`
`x=0.143=[OH^(-)]`
`[H^(+)]=(K_(w))/([OH^(-)])=7xx10^(-14)M`
`:.pH=13.15`